since y and u are both functions of x u can not apply the partial derivative (the chain rule formula)

What do you mean by partial derivative here??? You only encounter partial derivative in multi-variable function, not 1 variable function like f(x)!!!!

A_I_ said:

the chain rule is used when u have:

f(x,y) and x=g(t) and y=r(t)

but since your f and your u are both functions of x,
thus you can not use the chain rule and you can not say:

f'(x) = (dy/du)*(du/dx)

ok?

???
NOOO!!!!!! What do you mean by this??? I am TOTALLY lost!!!
Why can't you use the chain rule in that case? Am I misssing something?
Please look back at your cal textbook, see the part that covers the chain rule.
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[tex]\frac{df}{dx} = \frac{df}{du} \times \frac{du}{dx}[/tex]
Example:
[tex]\frac{df(2x)}{dx} = \frac{df(2x)}{d(2x)} \times \frac{d(2x)}{dx} = 2 f'(2x)[/tex].
Now, let's do your problem in a slightly different way:
[tex]\frac{df(2x)}{dx} = 2 f'(2x) = x ^ 2[/tex]
[tex]\Rightarrow f'(2x) = \frac{x ^ 2}{2}[/tex]
Now, let y = 2x, we have:
[tex]f'(y) = \frac{(2x) ^ 2}{8} = \frac{y ^ 2}{8}[/tex]
Now, what's f'(x)?
Can you go from here? Is there anything unclear? :)
-------------------
Whoops, looking back at some previous posts of this thread, I saw that you've looked through the manual.
The only error you made is that you were trying to find dy / dx, which means you were finding d(f(2x)) / dx, not d(f(x)) / dx (which means the same as f'(x)).
Is there anything unclear, endeavor? :)

for the first part, thats what i was trying to tell him,
that we only use the chain rule when we have a multivariable function, which is not the case.
As for the second part, i am not sure about it, because i know you can only use the chain rule with multivariable function.
Did u consider x and 2x to be two different variables, if yes, then it works.

for the first part, thats what i was trying to tell him,
that we only use the chain rule when we have a multivariable function, which is not the case.
As for the second part, i am not sure about it, because i know you can only use the chain rule with multivariable function.
Did u consider x and 2x to be two different variables, if yes, then it works.

What do you mean???Chain rule can be used for bothmultivariable functions, and 1 variable function.
You must have studied [tex]\frac{df}{dx} = \frac{df}{du} \times \frac{du}{dx}[/tex] BEFORE studying multi-variable functions, no?
As I told you before, you should re-read your calculus textbook, just look up the chapter for Chain rule (or you can just click on the link) for one variable function. That won't do you any harm, I promise.

So when in my first post I wrote:
[tex]\frac{dy}{du}[/tex]
that really means
[tex]\frac{d f(2x)}{du}[/tex]
since y = f(2x) ?

Yes, this is correct. :)

and then later on, when I wrote:
[tex]\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}[/tex]
I was actually finding f'(2x), which was already given??

Nah, this is not correct, you are finding:
[tex]\frac{df(2x)}{dx}[/tex], not f'(2x).
[tex]\frac{df(2x)}{dx} \neq f'(2x)[/tex].
To find f'(2x), you must find: [tex]\frac{df(2x)}{d(2x)}[/tex].
Can you get this? :)